mohan changed 2 in to 5 p 20 p and 50 p coins. if the number of 5 p coins was double the number of 20 p coins and the 50 p coins were one fifth the 20 p coins . how many coins of each denomination did he get?
Mohan changed 2 into 5p, 20p and 50p coins. If the number of 5p coins was double the number 20p coins, and the 50p coins were one-fifth the 20p coins, how many coins of each denomination di
Let's solve this step by step:
Step 1: Let's assume the number of 20 p coins as 'x'.
Step 2: According to the given problem, the number of 5 p coins is double the number of 20 p coins, so it will be '2x'.
Step 3: Also, the number of 50 p coins is one-fifth of the number of 20 p coins, so it will be 'x/5'.
Step 4: Now, we need to add up the total value of all the coins: 2x+20x+50(x/5) = 100
Step 5: Simplifying the equation: 2x + 20x + 10x = 100
Step 6: Combining the like terms: 32x = 100
Step 7: Dividing both sides of the equation by 32: x = 100/32
Step 8: Evaluating x: x = 3.125
Step 9: Since the number of coins cannot be in decimal, we can round x to the nearest whole number. Therefore, x ≈ 3.
So, Mohan will get approximately 3 coins of 20 p, 2 times 3 (i.e., 6) coins of 5 p, and 3/5 of 3 (i.e., 1.8, rounded to 2) coins of 50 p.
In conclusion, Mohan will get 3 coins of 20 p, 6 coins of 5 p, and 2 coins of 50 p.
To solve this problem, we can use a system of equations. Let's assign variables to represent the number of coins in each denomination:
Let's say the number of 5 pence coins is x
The number of 20 pence coins is y
The number of 50 pence coins is z
From the given information, we have three equations:
1. The number of 5 pence coins is double the number of 20 pence coins:
x = 2y
2. The number of 50 pence coins is one-fifth the number of 20 pence coins:
z = (1/5)y
3. The total value of all the coins is 2 pounds, which can also be represented as 200 pence:
5x + 20y + 50z = 200
Now, we can substitute equations 1 and 2 into equation 3 and solve the resulting system of equations.
Substituting the value of x from equation 1 into equation 3:
5(2y) + 20y + 50z = 200
10y + 20y + 50z = 200
30y + 50z = 200 (equation 4)
Substituting the value of z from equation 2 into equation 4:
30y + 50(1/5)y = 200
30y + 10y = 200
40y = 200
y = 5
Now we know that y = 5. Substituting this value into equations 1 and 2, we can find the values of x and z:
From equation 1:
x = 2y = 2 * 5 = 10
So, x = 10.
From equation 2:
z = (1/5)y = (1/5) * 5 = 1
So, z = 1.
Therefore, Mohan got 10 coins of 5 pence, 5 coins of 20 pence, and 1 coin of 50 pence.
If there were x,y,z 5p,20p, and 50p coins, respectively, then
x = 2y
y = 5z
5x+20y+50z = 200
Now just find x,y,z.